Question

If \[\left| \alpha \right|<1\] and \[y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+.....\infty \], then the value of \[\alpha \]is
\[\begin{align}
  & (\text{A) y+}\dfrac{1}{y} \\
 & (B)\text{ }\dfrac{y}{1-y} \\
 & (C)\text{ y-}\dfrac{1}{y} \\
 & (D)\text{ }\dfrac{y}{1+y} \\
\end{align}\]

Answer 1

Hint: We should also know that if a is the first term and b is second term, then \[r=\dfrac{b}{a}\]. In y, the first term is equal to \[\alpha \] and the second term is equal to \[-{{\alpha }^{2}}\]. Let us assume the common ratio in y is equal to r. Now we have to find the value of r. We should know that if the sum of infinite G.P is equal to S whose first term is a and common ratio is r, then\[S=a+ar+a{{r}^{2}}+a{{r}^{3}}+.......\infty =\dfrac{a}{1-r}\] where \[\left| r \right|<1\]. Now we have to find the value of y using the infinite G.P formula.

Complete step-by-step answer:
Before solving the question, we should know that if the sum of infinite G.P is equal to S whose first term is a and common ratio is r, then\[S=a+ar+a{{r}^{2}}+a{{r}^{3}}+.......\infty =\dfrac{a}{1-r}\] where \[\left| r \right|<1\]. We should also know that if a is the first term and b is second term, then \[r=\dfrac{b}{a}\].
In y, the first term is equal to \[\alpha \] and the second term is equal to \[-{{\alpha }^{2}}\]. Let us assume the common ratio in y is equal to r.
\[\begin{align}
  & \Rightarrow r=\dfrac{-{{\alpha }^{2}}}{\alpha } \\
 & \Rightarrow r=\alpha ....(1) \\
\end{align}\]
So, the common ratio of y is equal to \[\-alpha \].
We know that if the sum of infinite G.P is equal to S whose first term is a and common ratio is r, then\[S=a+ar+a{{r}^{2}}+a{{r}^{3}}+.......\infty =\dfrac{a}{1-r}\] where \[\left| r \right|<1\].
From the question, it was given that \[\left| \alpha \right|<1\] and \[y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+.....\infty \].
So, now we have to find the value of y where \[y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+.....\infty \] and \[\left| \alpha \right|<1\].
\[\begin{align}
  & \Rightarrow y=\dfrac{\alpha }{1-(-\alpha )} \\
 & \Rightarrow y=\dfrac{\alpha }{1+\alpha }....(2) \\
\end{align}\]
Now by using cross multiplication, we get
\[\begin{align}
  & \Rightarrow y+y\alpha =\alpha \\
 & \Rightarrow \alpha -y\alpha =y \\
 & \Rightarrow \alpha (1-y)=y \\
\end{align}\]
Now by using cross multiplication, we get
\[\alpha =\dfrac{y}{1-y}....(3)\]
From equation (3), it is clear that the value of \[\alpha \]is equal to \[\dfrac{y}{1-y}\].
Hence, option B is correct.

Note: Students may go wrong while assuming the value of common ratio. If students assume that the value of r is equal to \[\alpha \], then we get
\[\begin{align}
  & \Rightarrow y=\dfrac{\alpha }{1-\alpha } \\
 & \Rightarrow y-y\alpha =\alpha \\
 & \Rightarrow y=y\alpha +\alpha \\
 & \Rightarrow y=\alpha (1+y) \\
 & \Rightarrow \alpha =\dfrac{y}{1+y} \\
\end{align}\]
Hence, we get option D as correct. But we know that this option is incorrect.